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Available online at www.tjnsa.com J. Nonlinear Sci. Appl. 9 (2016), 4003–4015 Research Article On the split equality common fixed point problem for asymptotically nonexpansive semigroups in Banach spaces Zhaoli Ma a , Lin Wang b,* a School of Information Engineering, The College of Arts and Sciences Yunnan Normal University, Kunming, Yunnan, 650222, China. b College of Statistics and Mathematics, Yunnan University of Finance and Economics, Kunming, Yunnan 650221, China. Communicated by Y. J. Cho Abstract In this article, we propose an iteration methods for finding a split equality common fixed point of asymptotically nonexpansive semigroups in Banach spaces. The weak and strong convergence theorems of the iteration scheme proposed are obtained. As application, we shall utilize our results to study the split equality variational inequality problems to support the main results. The results presented in the article are new and improve and extend some recent corresponding results. c 2016 All rights reserved. Keywords: Split equality problem, convergence, asymptotically nonexpansive semigroup, Banach spaces. 2010 MSC: 47H09, 47H05, 47J20. 1. Introduction Let H 1 and H 2 be two real Hilbert spaces, C and Q be nonempty closed convex subsets of H 1 and H 2 , respectively. The split feasibility problem is formulated as finding a point q H 1 with the property: q C and Aq Q, (1.1) where A : H 1 H 2 is a bounded linear operator. Assuming that SFP (1.1) is consistent (that is, (1.1) has a solution), it is not hard to see that x C is a solution of (1.1) if and only if it solves the following fixed point equation x = P C (I - γA * (I - P Q )A)x, x C, * Corresponding author:Lin Wang Email addresses: [email protected] (Zhaoli Ma), [email protected] and [email protected] (Lin Wang) Received 2016-05-17

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Page 1: On the split equality common xed point problem for ... · On the split equality common xed point problem for asymptotically nonexpansive semigroups in Banach spaces Zhaoli Maa, Lin

Available online at www.tjnsa.comJ. Nonlinear Sci. Appl. 9 (2016), 4003–4015

Research Article

On the split equality common fixed point problemfor asymptotically nonexpansive semigroups inBanach spaces

Zhaoli Maa, Lin Wangb,∗

aSchool of Information Engineering, The College of Arts and Sciences Yunnan Normal University, Kunming, Yunnan, 650222, China.bCollege of Statistics and Mathematics, Yunnan University of Finance and Economics, Kunming, Yunnan 650221, China.

Communicated by Y. J. Cho

Abstract

In this article, we propose an iteration methods for finding a split equality common fixed point ofasymptotically nonexpansive semigroups in Banach spaces. The weak and strong convergence theorems ofthe iteration scheme proposed are obtained. As application, we shall utilize our results to study the splitequality variational inequality problems to support the main results. The results presented in the articleare new and improve and extend some recent corresponding results. c©2016 All rights reserved.

Keywords: Split equality problem, convergence, asymptotically nonexpansive semigroup, Banach spaces.2010 MSC: 47H09, 47H05, 47J20.

1. Introduction

Let H1 and H2 be two real Hilbert spaces, C and Q be nonempty closed convex subsets of H1 and H2,respectively. The split feasibility problem is formulated as finding a point q ∈ H1 with the property:

q ∈ C and Aq ∈ Q, (1.1)

where A : H1 → H2 is a bounded linear operator. Assuming that SFP (1.1) is consistent (that is, (1.1) hasa solution), it is not hard to see that x ∈ C is a solution of (1.1) if and only if it solves the following fixedpoint equation

x = PC(I − γA∗(I − PQ)A)x, x ∈ C,

∗Corresponding author:Lin WangEmail addresses: [email protected] (Zhaoli Ma), [email protected] and [email protected] (Lin Wang)

Received 2016-05-17

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Z. L. Ma, L. Wang, J. Nonlinear Sci. Appl. 9 (2016), 4003–4015 4004

where PC and PQ are the (orthogonal) projections onto C and Q, respectively, γ > 0 is any positive constant,and A∗ denotes the adjoint of A.

The split feasibility problem in finite dimensional Hilbert spaces was introduced by Censor and Elfving [5]in 1994 for modeling inverse problems which arise from phase retrievals and in medical imagine reconstruction[3]. Recently, it has been found that split feasibility problems can be used in various disciplines, such asimagine restoration, computer tomography, and radiation therapy treatment planning [4, 6, 7]. As well asthe convex feasibility formalism is at the core of the modeling of many inverse problems and has been usedto model significant real-world problems.

If C and Q are the sets of fixed points of two nonlinear mappings, respectively, and C and Q are nonemptyclosed convex subsets, then q is said to be a split common fixed point for the two nonlinear mappings. Thatis, the split common fixed point problem (SCFP ) for mappings S and T is to find a point q ∈ H1 with theproperty:

q ∈ C := F (S) and Aq ∈ Q := F (T ), (1.2)

where F (S) and F (T ) denote the sets of fixed points of S and T , respectively. We use Γ to denote the setof solution of SCFP (1.2), that is, Γ = q ∈ F (S) : Aq ∈ F (T ).

Recently, Moudafi [12] proposed a new split feasibility problem, which is also called split equality fixedpoint problem. Let H1, H2, and H3 be real Hilbert spaces, let U : H1 → H1 and T : H2 → H2 betwo nonlinear mappings with nonempty fixed point sets C := FixU and Q := FixT , A : H1 → H3 andB : H2 → H3 be two bounded linear operators. The split equality fixed point problem for U and T is

Finding x ∈ C and y ∈ Q such that Ax = By, (1.3)

which allows asymmetric and partial relations between the variables x and y. The interest is to cover manysituations, for instance in decomposition methods for PDE′s, applications in game theory and in intensity-modulated radiation therapy (IMRT). In decision sciences, this allows to consider agents who interplay onlyvia some components of their decision variables, for further details, the interested reader is referred to [1].In IMRT, this amounts to envisage a weak coupling between the vector of doses absorbed in all voxels andthat of the radiation intensity, for further details, the interested reader is referred to [4, 6].

We use Γ to denote the set of solutions of the new split feasibility problem (1.3), that is,

Γ = (x, y) : Ax = By, x ∈ C, y ∈ Q. (1.4)

Let E be a real normed linear space and C be a nonempty closed convex subset of E. The mappingT : C → C is said to be nonexpansive if for all x, y ∈ C

‖Tx− Ty‖ ≤ ‖x− y‖.

The mapping T : C → C is said to be asymptotically nonexpansive if there exists a sequence kn ⊂[1,∞) with limn→∞ kn = 1 such that for all x, y ∈ C and each n ≥ 1

‖Tnx− Tny‖ ≤ kn‖x− y‖.

Being an important generalization of the class of nonexpansive mappings, the class of asymptoticallynonexpansive mappings was introduced by Goebel and Kirk [10] in 1972, who proved that if C is a nonemptyclosed convex subset of a real uniformly convex Banach space and T is an asymptotically nonexpansivemapping, then T has a fixed point.

Definition 1.1 ([19]). A one-parameter family F := T (t) : t ≥ 0 of E into itself is called a stronglycontinuous semigroup of Lipschitzian mappings on E if it satisfies the following conditions:

(i) T (0)x = x, for all x ∈ E;

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Z. L. Ma, L. Wang, J. Nonlinear Sci. Appl. 9 (2016), 4003–4015 4005

(ii) T (s+ t) = T (s)T (t), for all s, t ≥ 0;

(iii) for each x ∈ E, the mapping t 7→ T (t)x is continuous;

(iv) for each t > 0, there exists a bounded measurable function L(t) : [0,∞)→ [0,∞) such that

‖T (t)x− T (t)y‖ ≤ L(t)‖x− y‖, for all x, y ∈ E.

A strongly continuous semigroup of Lipschitzian mappings F is called strongly continuous semigroupof nonexpansive mappings if L(t) = 1 for all t > 0 and strongly continuous semigroup of asymptoticallynonexpansive if lim supt→∞ L(t) ≤ 1. Note that for asymptotically nonexpansive semigroup F , we can alwaysassume that L(t)t>0 is such that L(t) ≥ 1 for each t > 0, L(t) is nonincreasing in t, and limt→∞ L(t) = 1;otherwise we replace L(t), for each t > 0, with L(t) := maxsups≥tL(s), 1. We denote by F (F) the set ofall common fixed points of F , that is,

F (F) := x ∈ E : T (t)x = x, 0 ≤ t <∞ =⋂t≥0

F (T (t)).

If F satisfies (i)-(iii) and

lim supt→∞x∈D

‖T (t)x− T (s)T (t)x‖ = 0, for all s > 0 and bounded D ⊆ C,

then F is called uniformly asymptotically regular on C.

Example 1.2 ([14], Example of asymptotically nonexpansive semi-group). Let E be an uniformly convexBanach space which admits a weakly continuous duality mapping. Let L(E) be the space of all boundedlinear operators on E. For Ψ ∈ L(E), define F := T (t) : t ∈ R+ of bounded linear operators by using thefollowing exponential expression:

T (t) = e−tΨ :=∞∑k=0

(−1)k

k!tkΨk.

Then, clearly, the family F := T (t) : t ∈ R+ satisfies the semigroup properties. Moreover, this familyforms a one parameter semigroup of self-mappings of E because etΨ = [e−tΨ]−1 : E → E exists for eacht ∈ R+.

Concerning the weak and strong convergence of iterative sequences to approximate a solution of splitfeasibility problem and split equality problem have been studied by some authors in the setting of Hilbertspace (see, for example, [3, 6, 7, 9, 11, 13, 20] and the references therein). But according to the literature,we can find that there is no relevant literature about the convergence of the split feasibility problem and thesplit equality common fixed point problem for the operator semigroups in Banach spaces. Very recently, In2015, Takahashi and Yao [15] obtained some strong and weak convergence theorems by using hybrid methodsfor the split feasibility problem and split common null point problem in the setting of one Hilbert space andone Banach space. Then, Tang et al. [18] proved a weak convergence theorem and a strong convergencetheorem for split common fixed point problem involving a quasi-strict pseudo contractive mapping and anasymptotical nonexpansive mapping in the setting of two Banach spaces.

In this paper, motivated by the works above, we propose the following iterative algorithm to approximatea solution of the split equality fixed point problems of two asymptotically nonexpansive semigroups in thesetting of two Banach spaces. For any given x0 ∈ E1 and y0 ∈ E2, the sequence (xn, yn) is defined asfollows:

zn ∈ J3(Axn −Byn)

un = S(tn)(xn − γJ−11 A∗zn)

vn = T (tn)(yn + γJ−12 B∗zn)

yn+1 = βnvn + (1− βn)(yn + γJ−12 B∗zn)

xn+1 = βnun + (1− βn)(xn − γJ−11 A∗zn).

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Under some suitable conditions strong and weak convergence theorems are established. As application,we shall utilize our results to study the split equality variational inequality problem. The results presentedin this paper are new and improve and extend some recent corresponding results.

2. Preliminaries

We now recall some definitions and elementary facts which will be used in the proofs of our main results.Let E be a real Banach space with the dual E∗. The normalized duality mapping J from E to 2E

∗is

defined byJx = x∗ ∈ E∗ : 〈x, x∗〉 = ‖x‖2 = ‖x∗‖2, ∀x ∈ E,

where 〈·, ·〉 denotes the generalized duality pairing between E and E∗.

A Banach space E is said to be strictly convex if ‖x+y‖2 ≤ 1 for all x, y ∈ U = z ∈ E : ‖z‖ = 1 with

x 6= y. The modulus of convexity of E is defined by

δE(ε) = inf1− ‖1

2(x+ y)‖ : ‖x‖, ‖y‖ ≤ 1, ‖x− y‖ ≥ ε

for all ε ∈ [0, 2]. E is said to be uniformly convex if δE(0) = 0, and δE(ε) > 0 for all 0 < ε ≤ 2. A Hilbertspace is 2-uniformly convex, while Lp is maxp, 2−uniformly convex for every p > 1.

Let ρE : [0,∞)→ [0,∞) be the modulus of smoothness of E defined by

ρE(t) = sup1

2(‖x+ y‖+ ‖x− y‖)− 1 : x ∈ U, ‖y‖ ≤ t.

A Banach space E is said to be uniformly smooth if ρE(t)t → 0 as t→ 0. A typical example of uniformly

smooth Banach space is Lp, where p > 1. More precisely, Lp is minp, 2-uniformly smooth for every p > 1.Let q be a fixed real number with q > 1, then a Banach space E is said to be q−uniformly smooth if thereexists a constant c > 0 such that ρE(t) ≤ ctq for all t > 0. It is well known that every q−uniformly smoothBanach space is uniformly smooth.

Lemma 2.1 ([21]). Given a number r > 0. A real Banach space E is uniformly convex if and only if thereexists a continuous strictly increasing function g : [0,∞)→ [0,∞) with g(0) = 0 such that

‖tx+ (1− t)y‖2 ≤ t‖x‖2 + (1− t)‖y‖2 − t(1− t)g(‖x− y‖)

for all x, y ∈ E, t ∈ [0, 1], with ‖x‖ ≤ r and ‖y‖ ≤ r .

Let T : C → C be a mapping with F (T ) 6= ∅. Then T is said to be demiclosed at zero if for any xn ⊂ Cwith xn x and ‖xn − Txn‖ → 0, x = Tx.

A mapping T : C → C is said to be semi−compact, if for any sequence xn in C such that ‖xn−Txn‖ →0, (n→∞), there exists subsequence xnj of xn such that xxj converges strongly to x∗ ∈ C.

A Banach space E is said to satisfy Opial’s property if for any sequence xn in E, xn x, for anyy ∈ E with y 6= x, we have

lim infn→∞

‖xn − x‖ < lim infn→∞

‖xn − y‖.

Lemma 2.2 ([21]). Let E be a 2-uniformly smooth Banach space with the best smoothness constants K > 0.Then the following inequality holds:

‖x+ y‖2 ≤ ‖x‖2 + 2〈y, Jx〉+ 2‖Ky‖2, ∀x, y ∈ E.

Lemma 2.3 ([8]). Let E be a real uniformly convex Banach space, C be a nonempty closed subset of E,and let T : C → C be an asymptotically nonexpansive mapping. Then I − T is demiclosed at zero, that is,if xn ⊂ C converges weakly to a point p ∈ C and limn→∞ ‖xn − Txn‖ = 0, then p = Tp.

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Z. L. Ma, L. Wang, J. Nonlinear Sci. Appl. 9 (2016), 4003–4015 4007

Lemma 2.4 ([17]). Let an and bn be two nonnegative real number sequences and satisfy

an+1 ≤ (1 + bn)an, ∀n ≥ 1,

where an ≥ 0, bn ≥ 0 and∑∞

n=1 bn <∞. Then

(1) limn→∞ an exists;

(2) if lim infn→∞ an = 0, then limn→∞ an = 0.

3. Main results

Throughout this section, we assume that:

(1) Let E1 and E2 be real uniformly convex and 2-uniformly smooth Banach spaces satisfying Opial’scondition and with the best smoothness constant k satisfying 0 < k < 1√

2, E3 be a real Banach space.

(2) Let A : E1 → E3 and B : E2 → E3 be two bounded linear operators with adjoints A∗ and B∗,respectively.

(3) Let S(t) : t ≥ 0 : E1 → E1 be uniformly asymptotically regular asymptotically nonexpansivesemigroup with a bounded measurable function L(1)(t) : [0,∞)→ [1,∞) satisfying limt→∞ L

(1)(t) = 1and C :=

⋂t≥0 F (S(t)) 6= ∅, T (t) : t ≥ 0 : E2 → E2 be uniformly asymptotically regular family of

asymptotically nonexpansive semigroup with a bounded measurable function L(2)(t) : [0,∞)→ [1,∞)satisfying limt→∞L

(2)(t) = 1 and Q :=⋂t≥0 F (T (t)) 6= ∅, respectively.

Theorem 3.1. Let E1, E2, E3, A, B, C, Q, S(t) : t ≥ 0 and T (t) : t ≥ 0 be the same as above. Forany given (x0, y0) ∈ E1 × E2, the sequence (xn, yn) is generated by

zn ∈ J3(Axn −Byn)

un = S(tn)(xn − γJ−11 A∗zn)

vn = T (tn)(yn + γJ−12 B∗zn)

yn+1 = βnvn + (1− βn)(yn + γJ−12 B∗zn)

xn+1 = βnun + (1− βn)(xn − γJ−11 A∗zn),

(3.1)

where tn is a sequence of real numbers, βn is a sequence in (0, 1) and γ is a positive number satisfying

(1) tn > 0 and limn→∞ tn =∞;

(2) L(t) = maxL(1)(t), L(2)(t) and∑∞

n=1(L2(tn)− 1) <∞;

(3) lim infn→∞ βn(1− βn) > 0 and 1‖A‖2+‖B‖2 < γ < 2

‖A‖2+‖B‖2 .

If Γ = (x∗, y∗) ∈ E1 × E2 : Ax∗ = By∗, x∗ ∈ C, y∗ ∈ Q 6= ∅, then

(I) the sequence (xn, yn) converges weakly to a solution (x∗, y∗) ∈ Γ of (1.4).

(II) In addition, if there exists at least one S(t) ∈ S(t) : t ≥ 0 and one T (t) ∈ T (t) : t ≥ 0 aresemi-compact, respectively, then the sequence (xn, yn) converges strongly to a solution (x∗, y∗) ∈ Γof (1.4).

Proof. Now we prove the conclusion (I).It is well known that the normalized duality mapping J of a smooth, reflexive and strictly convex Banach

space is single-valued, one to one, and surjective. Since E1 and E2 are real uniformly convex and 2-uniformlysmooth Banach spaces, and E3 is a real Banach space, therefore, the iteration scheme of (3.1) is well defined.

We shall divide the proof into four steps.

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Step 1. We first show that limn→∞ Γn+1(x, y) exists.Setting en = xn − γJ−1

1 A∗zn and wn = yn + γJ−12 B∗zn. Let (x, y) ∈ Γ, it follows from Lemma 2.1 that

‖xn+1 − x‖2 = ‖(1− βn)en + βnun − x‖2

= ‖(1− βn)(en − x) + βn(un − x)‖2

≤ (1− βn)‖en − x‖2 + βn‖un − x‖2 − βn(1− βn)g1(‖en − un‖)= (1− βn)‖en − x‖2 + βn‖S(tn)en − x‖2 − βn(1− βn)g1(‖en − un‖)≤ (1− βn)‖en − x‖2 + βnL

2(tn)‖en − x‖2 − βn(1− βn)g1(‖en − un‖)= (1 + βn(L2(tn)− 1))‖en − x‖2 − βn(1− βn)g1(‖en − un‖).

(3.2)

Further, from Lemma 2.2, we have

‖en − x‖2 = ‖xn − γJ−11 A∗zn − x‖2

= ‖(xn − x)− γJ−11 A∗zn‖2

≤ ‖γJ−11 A∗J3zn‖2 + 2γ〈x− xn, J1J

−11 A∗zn〉+ 2k2‖x− xn‖2

≤ γ2‖A‖2‖zn‖2 + 2γ〈Ax−Axn, zn)〉+ 2k2‖xn − x‖2.

(3.3)

By (3.2) and (3.3), we have

‖xn+1 − x‖2 ≤ (1 + βn(L2(tn)− 1))[γ2‖A‖2‖zn‖2 + 2k2‖xn − x‖2

+ 2γ〈Ax−Axn, zn〉]− βn(1− βn)g1(‖zn − un‖)= (1 + βn(L2(tn)− 1))2k2‖xn − x‖2 + (1 + βn(L2(tn)− 1))[γ2‖A‖2‖zn‖2

+ 2γ〈Ax−Axn, zn)〉]− βn(1− βn)g1(‖en − un‖).

(3.4)

By using the similar argument as given above, we have

‖yn+1 − y‖2 ≤ (1 + βn(L2(tn)− 1))2k2‖yn − y‖2 + (1 + βn(L2(tn)− 1))[γ2‖B‖2‖zn‖2

+ 2γ〈Byn −By, zn〉]− βn(1− βn)g2(‖wn − vn‖).(3.5)

By adding (3.4) and (3.5), and by taking into account the fact that Ax = By and zn ∈ J3(Axn −Byn),we have

‖xn+1 − x‖2 + ‖yn+1 − y‖2 ≤ (1 + βn(L2(tn)− 1))2k2(‖x− xn‖2 + ‖y − yn‖2)

+ (1 + βn(L2(tn)− 1))γ2(‖A‖2 + ‖B‖2)‖zn‖2

− 2(1 + βn(L2(tn)− 1))γ〈Axn −Byn, zn〉− βn(1− βn)[g1(‖en − un‖) + g2(‖wn − vn‖)]≤ (1 + βn(L2(tn)− 1))2k2(‖x− xn‖2 + ‖y − yn‖2)

+ (1 + βn(L2(tn)− 1))γ[γ(‖A‖2 + ‖B‖2)‖Axn −Byn‖2

− 2‖Axn −Byn‖2]− βn(1− βn)[g1(‖en − un‖) + g2(‖wn − vn‖)]≤ (1 + βn(L2(tn)− 1))2k2(‖x− xn‖2 + ‖y − yn‖2)

− (1 + βn(L2(tn)− 1))γ[2− γ(‖A‖2 + ‖B‖2)]‖Axn −Byn‖2

− βn(1− βn)[g1(‖en − un‖) + g2(‖wn − vn‖)].

Setting Γn(x, y) := ‖xn − x‖2 + ‖yn − y‖2, we have

Γn+1(x, y) ≤ (1 + βn(L2(tn)− 1))2k2Γn(x, y)

− (1 + βn(L2(tn)− 1))γ[2− γ(‖A‖2 + ‖B‖2)]‖Axn −Byn‖2

− βn(1− βn)[g1(‖en − un‖) + g2(‖wn − vn‖)].(3.6)

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Since limtn→∞ L(tn) = 1,∑∞

n=1(L2(tn) − 1) < ∞, 0 < k < 1√2, 1‖A‖2+‖B‖2 < γ < 2

‖A‖2+‖B‖2 , so,

0 < 2− γ(‖A‖2 + ‖B‖2) < 1. It follows from (3.6) and Lemma 2.4 that the limn→∞ Γn+1(x, y) exists.

Step 2. We prove that limn→∞ ‖Axn −Byn‖ = 0, limn→∞ ‖xn − un‖ = 0, and limn→∞ ‖yn − vn‖ = 0.It follows from (3.6) that

(1 + βn(L2(tn)− 1))γ[2− γ(‖A‖2 + ‖B‖2]‖Axn −Byn‖2

+ βn(1− βn)[g(‖zn − un‖) + g(‖wn − un‖)]≤ (1 + βn(L2(tn)− 1))2k2Γn(x, y)− Γn+1(x, y)

≤ Γn(x, y) + βn(L2(tn)− 1))2k2Γn(x, y)− Γn+1(x, y).

Therefore, we obtain that

limn→∞

g1(‖en − un‖) = 0, limn→∞

g2(‖wn − vn‖) = 0,

andlimn→∞

‖Axn −Byn‖ = 0. (3.7)

By virtue of Lemma 2.1 and the properties of g1 and g2, we may get

limn→∞

‖en − un‖ = 0, (3.8)

andlimn→∞

‖wn − vn‖ = 0. (3.9)

Since‖xn − en‖ = ‖J1(xn − en)‖ = ‖γA∗J3(Axn −Byn)‖ ≤ γ‖A‖‖Axn −Byn‖,

and‖yn − wn‖ = ‖J2(yn − wn)‖ = ‖γB∗J3(Axn −Byn)‖ ≤ γ‖B‖‖Axn −Byn‖.

From (3.7), we may getlimn→∞

‖xn − en‖ = 0, (3.10)

andlimn→∞

‖yn − wn‖ = 0.

It follows from (3.8) and (3.10) that

limn→∞

‖xn − un‖ = 0. (3.11)

By (3.9) and (3.11), we havelimn→∞

‖yn − vn‖ = 0. (3.12)

Step 3. We prove that limn→∞ ‖xn − S(t)xn‖ = 0 and limn→∞ ‖yn − T (t)yn‖ = 0, for all t ∈ [0,+∞).It follows from (3.1) that

‖xn+1 − xn‖ = ‖βnun + (1− βn)(xn − γJ−11 A∗zn)− xn‖

= ‖βnun − βnxn − (1− βn)γJ−11 A∗zn)‖

≤ βn‖un − xn‖+ (1− βn)‖γJ−11 A∗J3(Axn −Byn))‖.

(3.13)

From (3.7), (3.11), and (3.13), we have

limn→∞

‖xn+1 − xn‖ = 0. (3.14)

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Z. L. Ma, L. Wang, J. Nonlinear Sci. Appl. 9 (2016), 4003–4015 4010

Similarly, we can obtainlimn→∞

‖yn+1 − yn‖ = 0. (3.15)

In addition, from (3.1) we can get

‖un − S(tn)xn‖2 = ‖S(tn)(xn − γJ−11 A∗zn)− S(tn)xn‖2

≤ L2(tn)‖γJ−11 A∗zn)‖2

≤ L2(tn)‖γJ−11 A∗J3(Axn −Byn)‖2,

and‖vn − T (tn)yn‖2 = ‖T (tn)(yn + γJ−1

2 B∗zn)− T (tn)yn‖2

≤ L2(tn)‖γJ−12 B∗J3(Axn −Byn)‖2.

By (3.7), we obtainlimn→∞

‖un − S(tn)xn‖ = 0, (3.16)

andlimn→∞

‖vn − T (tn)yn‖ = 0. (3.17)

It follows from (3.11), (3.12), (3.16), and (3.17) that

limn→∞

‖xn − S(tn)xn‖ = 0, (3.18)

andlimn→∞

‖yn − T (tn)yn‖ = 0.

Since ‖xn − x‖2 ≤ Γn(x, y), ‖yn − y‖2 ≤ Γn(x, y), and limn→∞ Γn exists, we know that xn and ynare bounded. Therefore, there exist bounded subsets C1 ⊆ E1 and Q1 ⊆ E2 such that xn ⊆ C1 andyn ⊆ Q1, respectively. Since S(t) : t ≥ 0 and T (t) : t ≥ 0 are uniformly asymptotically regular, andlimn→∞ tn =∞, then for all t ≥ 0,

limn→∞

‖S(t)S(tn)xn − S(tn)xn‖ ≤ lim supn→∞x∈C1

‖S(t)S(tn)x− S(tn)x‖ = 0,

andlimn→∞

‖T (t)T (tn)yn − T (t)yn‖ ≤ lim supn→∞y∈Q1

‖T (t)T (tn)Ay − T (t)Ay‖ = 0.

Since S(t)x is continuous on t for all x ∈ E1, and

‖xn − S(t)xn‖ ≤ ‖xn − S(tn)xn‖+ ‖S(tn)xn − S(t)S(tn)xn‖+ ‖S(t)S(tn)xn − S(t)xn‖. (3.19)

It follows from (3.18) and (3.19) that

limn→∞

‖xn − S(t)xn‖ = 0.

Similarly,limn→∞

‖yn − T (t)yn‖ = 0.

Step 4. We prove that (x∗, y∗) is the unique weak cluster point of (xn, yn).

Since E1 and E2 are uniformly convex, they are reflexive. On the other hand, since (xn, yn) ⊆ C1×Q1,so we may assume that (x∗, y∗) is a weak cluster point of (xn, yn). Since each asymptotically nonexpansivemapping on real uniformly convex Banach spaces is demiclosed at zero, we know from Lemma 2.3 thatx∗ ∈ C = ∩t≥0F (S(t)), y∗ ∈ Q = ∩t≥0F (T (t)).

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Since A and B are bounded linear operators, we know that Ax∗ − By∗ is a weak cluster point ofAxn −Byn. From the weakly lower semi-continuous property of the norm and (3.7), we get

‖Ax∗ −By∗‖ ≤ lim infn→∞

‖Axn −Byn‖ = 0.

So, Ax∗ = By∗. This implies (x∗, y∗) ∈ Γ.We now prove that (x∗, y∗) is the unique weak cluster point of (xn, yn).Assume that there exists another subsequence (xnk

, ynk) of (xn, yn) such that (xnk

, ynk) converges

weakly to a point (p, q) with (p, q) 6= (x∗, y∗). Similar with the argument above, we know that (p, q) ∈ Γ,too. Since E1 and E2 satisfy Opial’s property, we have

lim infi→∞

‖xni − p‖ < lim infi→∞

‖xni − x∗‖ = limn→∞

‖xn − x∗‖

= lim infk→∞

‖xnk− x∗‖ < lim inf

k→∞‖xnk

− p‖

= limn→∞

‖xn − p‖ = lim infi→∞

‖xni − p‖,

andlim infi→∞

‖yni − q‖ < lim infi→∞

‖yni − y∗‖ = limn→∞

‖yn − y∗‖

= lim infk→∞

‖ynk− y∗‖ < lim inf

k→∞‖ynk

− q‖

= limn→∞

‖yn − q‖ = lim infi→∞

‖yni − q‖,

which is a contradiction. This implies that (p, q) = (x∗, y∗). The proof of conclusion (I) is completed.Next, we prove the conclusion (II).Since there exist one S(t) ∈ S(t) : t ≥ 0 and one T (t) ∈ T (t) : t ≥ 0 are semi-compact, (xn, yn)

is bounded and limn→∞ ||xn − S(t)xn|| = 0, limn→∞ ||yn − T (t)yn|| = 0 for all t ≥ 0, then there existssubsequences (xnj , ynj ) of (xn, yn) such that (xnj , ynj ) converges strongly to (u∗, v∗). Due to (xn, yn)converges weakly to (x∗, y∗), we know that (u∗, v∗) = (x∗, y∗). It follows from Lemma 2.3 that (x∗, y∗) ∈C ×Q. Further, due to the norm ‖ · ‖ is weakly lower semi-continuous and Axnj −Bynj → Ax∗ −By∗, weget

‖Ax∗ −By∗‖2 ≤ lim infn→∞

‖Axnj −Bynj‖2 = 0.

So, Ax∗ = By∗. This implies (x∗, y∗) ∈ Γ.On the other hand, since Γn(x, y) = ‖xn − x‖2 + ‖yn − y‖2 for any (x, y) ∈ Γ, we know that

limj→∞ Γnj (x∗, y∗) = 0. From Conclusion (I), we know that limn→∞ Γn(x∗, y∗) exists, therefore,

limn→∞ Γn(x∗, y∗) = 0. From the facts that 0 ≤ ‖xn − x∗‖ ≤ Γn and 0 ≤ ‖yn − y∗‖ ≤ Γn, we can ob-tain that limn→∞ ‖xn − x∗‖ = 0 and limn→∞ ‖yn − y∗‖ = 0. This completes the proof of the Conclusion(II).

Corollary 3.2. Let E1, E2, E3, A, and B be the same as above. Let S(t) : t ≥ 0 : E1 → E1 andT (t) : t ≥ 0 : E2 → E2 be two nonexpansive semigroups satisfying C := ∩t≥0F (S(t)) 6= ∅ and Q :=∩t≥0F (T (t)) 6= ∅, respectively. For any given (x0, y0) ∈ E1 × E2, the sequence (xn, yn) is generated by

zn ∈ J3(Axn −Byn)

un = S(tn)(xn − γJ−11 A∗zn)

vn = T (tn)(yn + γJ−12 B∗zn)

yn+1 = βnvn + (1− βn)(yn + γJ−12 B∗zn)

xn+1 = βnun + (1− βn)(xn − γJ−11 A∗zn),

where tn is a sequence of real numbers, βn is a sequence in (0, 1) and γ is a positive real numbersatisfying

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(1) tn > 0 and limn→∞ tn =∞;

(2) lim infn→∞ βn(1− βn) > 0 and 1‖A‖2+‖B‖2 < γ < 2

‖A‖2+‖B‖2 .

If Γ = (x∗, y∗) ∈ E1 × E2 : Ax∗ = By∗, x∗ ∈ C, y∗ ∈ Q 6= ∅, then

(I) the sequence (xn, yn) converges weakly to a solution (x∗, y∗) ∈ Γ of (1.4).

(II) In addition, if there exists at least one S(t) ∈ S(t) : t ≥ 0 and one T (t) ∈ T (t) : t ≥ 0 aresemi-compact, respectively, then the sequence (xn, yn) converges strongly to a solution (x∗, y∗) ∈ Γof (1.4).

In Theorem 3.1, taking B = I, E2 = E3, J2 = J3, similar with the proofs in Theorem 3.1, we can obtainthe following result for split common fixed point problem (1.2).

Corollary 3.3. Let E1, A, S(t) : t ≥ 0 and T (t) : t ≥ 0 be the same as Theorem 3.1, E2 be a realBanach space. For any given (x0, y0) ∈ E1 × E2, the sequence (xn, yn) is generated by

zn ∈ J2(Axn − yn)

un = S(tn)(xn − γJ−11 A∗zn)

vn = T (tn)(yn + γ(Axn − yn))

yn+1 = βnvn + (1− βn)(yn + γ(Axn − yn))

xn+1 = βnun + (1− βn)(xn − γJ−11 A∗zn),

where tn is a sequence of real numbers, βn is a sequence in (0, 1) and γ is a positive real numbersatisfying

(1) tn > 0 and limn→∞ tn =∞;

(2) L(t) = maxL(1)(t), L(2)(t) and∑∞

n=1(L2(tn)− 1) <∞;

(3) lim infn→∞ βn(1− βn) > 0 and 1‖A‖2+1

< γ < 2‖A‖2+1

.

If Γ = p ∈ C : Ap ∈ Q 6= ∅, then

(I) the sequence (xn, yn) converges weakly to a solution (x∗, y∗) ∈ Γ of (1.2).

(II) In addition, if there exists at least one S(t) ∈ S(t) : t ≥ 0 and one T (t) ∈ T (t) : t ≥ 0 aresemi-compact, respectively, then the sequence (xn, yn) converges strongly to a solution (x∗, y∗) ∈ Γof (1.2).

Corollary 3.4. Let E1, E2, E3, A, and B be the same as Theorem 3.1, S : E1 → E1 and T : E2 → E2

be two asymptotically nonexpansive mapping with the sequence kn ⊆ [1,∞) and ln ⊆ [1,∞) satisfying∑∞n=1(kn−1) < +∞, and

∑∞n=1(ln−1) < +∞, respectively. For any given (x0, y0) ∈ E1×E2, the sequence

(xn, yn) is generated by

zn ∈ J3(Axn −Byn)

un = Sn(xn − γJ−11 A∗zn)

vn = Tn(yn + γJ−12 B∗zn)

yn+1 = βnvn + (1− βn)(yn + γJ−12 B∗zn)

xn+1 = βnun + (1− βn)(xn − γJ−11 A∗zn),

where βn is a sequence in (0, 1) and lim infn→∞ βn(1−βn) > 0,γ is a positive number satisfying 1‖A‖2+‖B‖2 <

γ < 2‖A‖2+‖B‖2 . If Γ = (x∗, y∗) ∈ E1 × E2 : Ax∗ = By∗, x∗ ∈ C, y∗ ∈ Q 6= ∅, where C := F (S) 6= ∅ and

Q := F (T ) 6= ∅, then

(I) the sequence (xn, yn) converges weakly to a solution (x∗, y∗) ∈ Γ of (1.4).

(II) In addition, if S and T are semi-compact, then the sequence (xn, yn) converges strongly to a solution(x∗, y∗) ∈ Γ of (1.4).

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4. Application to the split equality variational inequality problem in Banach spaces

Throughout this section, we assume that C and Q are nonempty and closed convex subsets of E1 andE2, respectively.

Let M : C → E1 be a mapping. Variational inequality problem (VIP) in Banach space is formulated asthe problem of finding a point x∗ with property x∗ ∈ C such that for some j(z − x∗) ∈ J(x− x∗),

〈Mx∗, j(z − x∗)〉 ≥ 0, ∀z ∈ C.

We will denote the solution set of VIP by VI(M, C).A mapping M : C → E1 is said to be α-strongly accretive if for any x, y ∈ C, there exists j(x − y) ∈

J(x− y) such that〈Mx−My, j(x− y)〉 ≥ α‖x− y‖2, for α > 0.

A mapping M : C → E1 is said to be β-inverse strongly accretive if for each x, y ∈ C, there existsj(x− y) ∈ J(x− y) such that

〈Mx−My, j(x− y)〉 ≥ β‖Mx−My‖2, for β > 0.

The equilibrium problem (for short, EP ) is to find x∗ ∈ C such that

F (x∗, y) ≥ 0, ∀y ∈ C.

The set of solutions of EP is denoted by EP (F ). Given a mapping T : C → C, let F (x, y) =<Tx, j(y − x) > for all x, y ∈ C. Then x∗ ∈ EP (F ) if and only if x∗ ∈ C is a solution of the variationalinequality < Tx, j(y − x) >≥ 0 for all y ∈ C, that is, x∗ is a solution of the variational inequality.

Setting F (x, y) = 〈Mx, j(y − x)〉, it is easy to show that F satisfies the following conditions (A1)–(A4)as M is a β-inverse strongly accretive mapping.

(A1) F (x, x) = 0, ∀x ∈ C;

(A2) F (x, y) + F (y, x) ≤ 0, ∀x, y ∈ C;

(A3) For all x, y, z ∈ C, limt↓0 F (tz + (1− t)x, y) ≤ F (x, y);

(A4) For each x ∈ C, the function y 7−→ F (x, y) is convex and lower semi-continuous.

Lemma 4.1 ([2]). Let C be a closed convex subset of a reflexive, strictly convex and smooth Banach spaceE. Suppose F is a bifunction from C × C to R satisfying (A1)–(A4), r > 0 and x ∈ E. Then, there existsz ∈ C such that

F (z, y) +1

r〈y − z, j(z − x)〉 ≥ 0, ∀y ∈ C.

Lemma 4.2 ([16]). Let C be a closed convex subset of a uniformly smooth, strictly convex and reflexiveBanach space E. Suppose F is a bifunction from C × C to R satisfying (A1)–(A4). For r > 0 and x ∈ E,define a mapping Tr : X → C as follows:

Tr(x) = z ∈ C : F (z, y) +1

r< y − z, j(z − x) >≥ 0, ∀y ∈ C.

Then,

(1) Tr is single-valued;

(2) Tr is firmly nonexpansive, that is, ∀x, y ∈ E,

〈Tr(x)− Tr(y), j(Tr(x))− j(Tr(y))〉 ≤< Tr(x)− Tr(y), j(x)− j(y) >;

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Z. L. Ma, L. Wang, J. Nonlinear Sci. Appl. 9 (2016), 4003–4015 4014

(3) F (Tr) = EP (F );

(4) EP (F ) is closed and convex and Tr is a relatively nonexpansive mappings.

Let B1 : C → E1 and B2 : Q → E2 be two β-inverse-strongly accretive mappings, where C and Q arenonempty and closed convex subsets of E1 and E2, respectively. The ”so-called” split equality variationalinequality problem is shown that it is equivalent to find x∗ ∈ C, y∗ ∈ Q such that

< B1(x∗), j1(x− x∗) >≥ 0, for all x ∈ C,

and< B2(y∗), j2(y − y∗) >≥ 0, for all y ∈ Q,

and such thatAx∗ = By∗. (4.1)

We will denote the solution set of split equality variational inequality problem by Ω, that is,

Ω = x∗, y∗) ∈ V I(B1, C)× V I(B2, Q) : Ax∗ = By∗.

Setting F (x, y) =< B1x, j1(y − x) >, and G(x, y) =< B2x, j2(y − x) >, it is easy to show that F andG satisfy the conditions (A1)–(A4) as Bi (i = 1, 2) is a βi−inverse-strongly accretive mapping. For r > 0,x ∈ E1 and u ∈ E2, define mappings Tr : E1 → C and Sr : E2 → Q as follows:

Tr(x) = z ∈ C : F (z, y) +1

r< y − z, j1(z − x) >≥ 0, ∀y ∈ C,

and

Sr(u) = z ∈ Q : G(z, v) +1

r< v − z, j2(z − u) >≥ 0, ∀v ∈ Q.

It follows from Lemma 4.2 that F (Tr) = V I(B1, C) 6= ∅, F (Sr) = V I(B2, Q) 6= ∅, Tr(x) and Sr(u)are single-valued and firmly nonexpansive mappings, respectively. Therefore the split equality variationalinequality problem with respect to B1 and B2 is equivalent to the following split equality fixed point problem:

to find x∗ ∈ F (Tr), y∗ ∈ F (Sr) such that Ax

∗ = By∗.

Then it follows from Theorem 3.1 that the following result holds.

Theorem 4.3. Let E1 and E2 be real uniformly convex and 2-uniformly smooth Banach spaces satisfyingOpial’s condition and with the best smoothness constant k satisfying 0 < k < 1√

2, E3 be a real Banach space,

C ⊆ E1 and Q ⊆ E2 be nonempty closed convex subsets of E1 and E2, respectively. Let Bi (i = 1, 2) is aηi−inverse strongly accretive mappings, and A : E1 → E3 and B : E2 → E3 be two bounded linear operatorswith adjoints A∗ and B∗, respectively, Tr and Sr be the resolvent operator of the equilibrium function F andG, respectively. For any given (x0, y0) ∈ E1 × E2, the sequence (xn, yn) is generated by

zn ∈ J3(Axn −Byn)

un = Tr(xn − γJ−11 A∗zn)

vn = Sr(yn + γJ−12 B∗zn)

yn+1 = βnvn + (1− βn)(yn + γJ−12 B∗zn)

xn+1 = βnun + (1− βn)(xn − γJ−11 A∗zn),

where βn is a sequence in (0, 1) and lim infn→∞ βn(1 − βn) > 0, γ is a positive real number satisfying1

‖A‖2+‖B‖2 < γ < 2‖A‖2+‖B‖2 . If Ω = x∗, y∗) ∈ V I(B1, C)× V I(B2, Q) : Ax∗ = By∗ 6= ∅, then

(I) the sequence (xn, yn) converges weakly to a solution (x∗, y∗) ∈ Ω of (4.1);

(II) In addition, if Sr and Tr are semi-compact, then the sequence (xn, yn) converges strongly to a solution(x∗, y∗) ∈ Ω of (4.1).

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Acknowledgment

The authors wish to thank the editor and the referee for valuable suggestions. This work was supportedby the National Natural Science Foundation of China (Grant No. 11361070).

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